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Chapter 4: Problem 108

A seawater sample has a density of \(1.03 \mathrm{g} / \mathrm{mL}\) and \(2.8 \% \mathrm{NaCl}\)l by mass. A saturated solution of \(\mathrm{NaCl}\) in water is \(5.45 \mathrm{M} \mathrm{NaCl} .\) How many liters of water would have to be evaporated from \(1.00 \times 10^{6} \mathrm{L}\) of the seawater before \(\mathrm{NaCl}\) would begin to crystallize? (A saturated solution contains the maximum amount of dissolved solute possible.)

Short Answer

Expert verified
Approximately \(9.29 \times 10^4 \mathrm{L}\) of water would have to be evaporated from the seawater for NaCl to start crystallizing.

Step by step solution

01

Calculate the total mass of the seawater

The given density of the seawater is \(1.03 \mathrm{g}/\mathrm{mL}\) or \(1030 \mathrm{kg}/\mathrm{m}^3\), and the volume is \(1.00 \times 10^{6} \mathrm{L}\) or \(1000 \mathrm{m}^3\). Using the formula: Density = Mass / Volume, the total mass of the seawater is \(1.03 \times 10^6 \mathrm{kg}\).
02

Calculate the mass of NaCl in seawater

The percentage of NaCl by mass is 2.8%. Therefore, the mass of NaCl in the seawater can be calculated as: \(2.8/100 \times 1.03 \times 10^6 = 28840 \mathrm{kg}\).
03

Calculate the moles of NaCl in seawater

We know that the molar mass of NaCl is approximately \(58.44 \mathrm{g/mol}\) or \(0.05844 \mathrm{kg/mol}\). The number of moles of NaCl in the seawater is thus: \(28840/0.05844 = 4.94 \times 10^5 \mathrm{moles}\).
04

Calculate the amount of water to be evaporated

A saturate solution of NaCl is \(5.45 \mathrm{M}\), which means there are \(5.45 \mathrm{moles}\) of NaCl in 1 liter of water. The amount of water in which \(4.94 \times 10^5 \mathrm{moles}\) of NaCl can be dissolved is \((4.94 \times 10^5)/5.45 = 9.07 \times 10^4 \mathrm{L}\). Since the total volume of the seawater is \(1.00 \times 10^6 \mathrm{L}\), the amount of water that needs to be evaporated is \(1.00 \times 10^6 - 9.07 \times 10^4 = 9.29 \times 10^4 \mathrm{L}\).

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Most popular questions from this chapter

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